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a body weighing 500n sitting on a see saw at a point 2m away from pivot point.what should be the force acting on the other side of the pivot point 4m away to balance? 3000N 1000N 250N 750N

Physics
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f1 = 500n r1 = 2m r2 = 4 m f2 = ? Applying the second condition of equilibrium..... f1*r1=f2*r2, f2 =(f1*r1)*r2=(500n*2m)/4m =250n,,,, Third choice is right 250N

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